text: Here is a sample molecule with the parent carbons numbered:
  For simplicity, here is an image of the same molecule, where the hydrogens in the parent chain are removed and the carbons are shown by their numbers:
  Now, following the above steps:
  # The parent hydrocarbon chain has 23 carbons. It is called tricosa-.
  # The functional groups with the highest precedence are the two ketone groups.
  ## The groups are on carbon atoms 3 and 9. As there are two, we write 3,9-dione.
  ## The numbering of the molecule is based on the ketone groups. When numbering from left to right, the ketone groups are numbered 3 and 9. When numbering from right to left, the ketone groups are numbered 15 and 21. 3 is less than 15, therefore the ketones are numbered 3 and 9. The smaller number is always used, not the sum of the constituents numbers.
  # The side chains are: an ethyl- at carbon 4, an ethyl- at carbon 8, and a butyl- at carbon 12. <br/>Note: the  at carbon atom 15 is not a side chain, but it is a methoxy functional group.
  #* There are two ethyl- groups. They are combined to create, 4,8-diethyl.
  #* The side chains are grouped like this: 12-butyl-4,8-diethyl. (But this is not necessarily the final grouping, as functional groups may be added in between to ensure all groups are listed alphabetically.)
  # The secondary functional groups are: a hydroxy- at carbon 5, a chloro- at carbon 11, a methoxy- at carbon 15, and a bromo- at carbon 18. Grouped with the side chains, this gives 18-bromo-12-butyl-11-chloro-4,8-diethyl-5-hydroxy-15-methoxy.
  # There are two double bonds: one between carbons 6 and 7, and one between carbons 13 and 14. They would be called "6,13-diene", but the presence of alkynes switches it to 6,13-dien. There is one triple bond between carbon atoms 19 and 20. It will be called 19-yne.
  # The arrangement (with punctuation) is: 18-bromo-12-butyl-11-chloro-4,8-diethyl-5-hydroxy-15-methoxytricosa-6,13-dien-19-yne-3,9-dione
  # Finally, due to cis-trans isomerism, we have to specify the relative orientation of functional groups around each double bond. For this example, both double bonds are trans isomers, so we have (6E,13E)
  The final name is (6E,13E)-18-bromo-12-butyl-11-chloro-4,8-diethyl-5-hydroxy-15-methoxytricosa-6,13-dien-19-yne-3,9-dione.